Video Transcript
A 650-newton constant force is applied by a traveler in a winter landscape to pull a sled of mass 125 kilograms a distance of 1.4 meters across a frozen lake. While the sled is being dragged, it accelerates at an average rate of 0.12 meters per second squared. How much work is done in total moving the sled across the lake?
The question is about a traveler pulling a sled across a lake. So let’s begin by drawing a diagram that shows what’s going on. Okay, so here’s our traveler walking across the lake. And we’ve drawn them holding onto a rope such that they’re pulling a sled across this lake. The question tells us that the traveler applies a constant force of 650 newtons in order to pull the sled. Assuming that this force is entirely horizontal, let’s add it to our diagram, and we’ll label it as 𝐹.
We’re also told that the sled has a mass of 125 kilograms. Let’s label this mass as 𝑚. The traveler pulls this sled for a distance of 1.4 meters, which we can add to our diagram and label as 𝑑. The last bit of information given to us is that while the sled is being dragged, it accelerates at an average rate of 0.12 meters per second squared. We’ll label this acceleration as 𝑎, and we know that it will be in the same direction as the traveler is pulling the sled.
This first part of the question is asking us to find the total amount of work done in moving the sled across the lake. The total work done is all of the work that is done by the force applied by the traveler to move the sled. And we can recall that the work done on an object by a force is equal to the magnitude of the force multiplied by the distance that the object moves. So if we label the total work done in moving the sled as 𝑤, then this is equal to the force 𝐹 that the traveler applies multiplied by the distance 𝑑 that they pull the sled. Then, in terms of symbols, we have that 𝑤, the work done, is equal to the force 𝐹 multiplied by the distance 𝑑.
So let’s take our values for the quantities 𝐹 and 𝑑 and sub them into this equation to calculate the total work done. When we do this, we find that 𝑤 is equal to 650 newtons, that’s our value for the force 𝐹, multiplied by 1.4 meters, that’s the distance 𝑑 that the sled moves. For a force in units of newtons and a distance in units of meters, we’ll calculate a work done in the SI unit of energy, the joule.
Evaluating the expression gives a result of 910 joules. So our answer for the first part of the question is that the total work done in moving the sled across the lake is 910 joules.
Okay, let’s now clear some space and look at the second part of the question.
How much work is done to accelerate the sled?
So the first part of the question was asking us for the total work done to move the sled. And now this second part of the question is asking us to find the work done to accelerate the sled. We might wonder why the second question is actually a different thing to this first one. The reason is that of this total work done, not all of it goes to accelerating the sled because some of the work will be used to overcome things such as friction and air resistance.
The force 𝐹 in this diagram is the constant force applied by the traveler. In order to find the work done to accelerate the sled, rather than using this force applied by the traveler, we need to find the value of the forward force that’s accelerating the sled. As we’ve said, because of things such as friction and air resistance, we expect that this force will be smaller than the force 𝐹 applied by the traveler. We can calculate this force using the mass 𝑚 of the sled and its average acceleration 𝑎.
In order to do this, we can recall Newton’s second law of motion, which tells us that force is equal to mass multiplied by acceleration. So in this case, the overall forward force on the sled, which we’ll label 𝐹 subscript 𝑠, must be equal to the mass 𝑚 of the sled multiplied by its acceleration 𝑎. That gives us this equation here.
We can now sub in our values for the mass 𝑚 and the acceleration 𝑎. When we do this, we get an equation that says 𝐹 subscript 𝑠 is equal to 125 kilograms, that’s the mass 𝑚, multiplied by 0.12 meters per second squared, that’s the acceleration 𝑎. A mass in units of kilograms and an acceleration in meters per second squared will give us a force 𝐹 subscript 𝑠 in the SI unit of force, the newton. Evaluating the expression, we get a result for the force on the sled, 𝐹 subscript 𝑠, of 15 newtons.
As in the first part of the question, we can now recall that work done is equal to force multiplied by distance. This time, our equation is that the work done to accelerate the sled, 𝑊 subscript 𝑠, is equal to the force that’s accelerating the sled, 𝐹 subscript 𝑠, multiplied by the distance 𝑑 that the sled moves.
Let’s take our values for 𝐹 subscript 𝑠 and 𝑑 and sub them into this equation. This gives us that 𝑊 subscript 𝑠 is equal to 15 newtons, our value for the force 𝐹 subscript 𝑠, multiplied by 1.4 meters, that’s the distance 𝑑. Evaluating this expression gives us a result of 21 joules. This means our answer to the second part of the question is that the amount of work done to accelerate the sled is 21 joules. So of the total of 910 joules of work that the traveler does to move the sled, only 21 joules of this goes towards accelerating the sled.
Okay, let’s now clear some space again, and we’ll look at the final part of the question.
The sled was at rest before the traveler started to pull it across the lake. What is the velocity of the sled just after it has crossed the lake? Round your answer to two decimal places.
Okay, so we know from the main text of the question that the sled is accelerating as it’s pulled across the lake. This last bit of the question is asking us to find the final velocity that it reaches after it has crossed the lake. Let’s label this final velocity as 𝑣. We’re told that before the traveler starts to pull it, the sled was at rest. This means that the sled had an initial velocity, which we’ve labeled as 𝑢, of zero meters per second. We know that the acceleration of the sled 𝑎 is equal to 0.12 meters per second squared and that it accelerates over a distance 𝑑 equal to 1.4 meters.
We can recall that there is a kinematic equation we can make use of which links an object’s initial velocity, its final velocity, its acceleration, and the distance it accelerates over. We usually see this equation written as 𝑣 squared is equal to 𝑢 squared plus two times 𝑎 times 𝑠, where 𝑣 is the object’s final velocity, 𝑢 is its initial velocity, 𝑎 is its acceleration, and 𝑠 is the distance over which it accelerates.
In this question, we’ve already labeled the distance as 𝑑 rather than 𝑠. So with this notation, our kinematic equation becomes 𝑣 squared is equal to 𝑢 squared plus two times 𝑎 times 𝑑. We can now take our values for the initial velocity 𝑢, the acceleration 𝑎, and the distance 𝑑 and sub them into this equation.
When we do this, we get an equation that says the final velocity 𝑣 squared is equal to the square of zero meters per second plus two times 0.12 meters per second squared times 1.4 meters. Evaluating this expression, we find that the square of zero meters per second is zero meters squared per second squared. And this second term, two times 0.12 meters per second squared times 1.4 meters, works out as 0.336 meters squared per second squared. And so we find that the square of the final velocity is equal to 0.336 meters squared per second squared.
To find the final velocity itself, all we need to do now is to take the square root of this value. Let’s clear ourselves a little bit of space to do this. Okay, so we’ve got our value for 𝑣 squared, and we know that 𝑣 must be equal to the square root of this value. So 𝑣 is equal to the square root of 0.336 meters squared per second squared. Evaluating the square root, we find that the final speed 𝑣 is equal to 0.5797 meters per second, where the ellipses indicate that there are further decimal places.
The question asks for our answer to be rounded to two decimal places. So taking this value and rounding it to two decimal places gives us our answer to the final part of the question that the velocity of the sled just after it has crossed the lake is equal to 0.58 meters per second.
